I have to calculate maximum of every unique pair of elements in matrix. So here is my code:
resultsMat = [
6 4 4;
0 2 6;
7 7 1;
5 1 73
];
copyMat = resultsMat;
for i=1:size(resultsMat,1)
for j=1:size(resultsMat,2)
for q=1:size(resultsMat,1)
for p=1:size(resultsMat,2)
if i== q && j ~= p
a = max(resultsMat(i,j),copyMat(q,p))
end
end
end
end
end
The problem comes when I try to store values in a matrix. For example:
[val ind] = max(resultsMat(i,j),copyMat(q,p))
This throws an error:
Error using max
MAX with two matrices to compare and two output arguments is not supported.
Error in Untitled2 (line 18)
[a, b] = max(resultsMat(i,j),copyMat(q,p))
How to store values from a = max(resultsMat(i,j),copyMat(q,p)) in a matrix?
You need a larger (probably multi-dimensional) matrix, as every (i,j) location has a maximum vs any (q,p) location. This means that for every element in your first matrix, you obtain a full matrix of the same size. Saving as
matrix_with_results(i,j,q,p) = a
would do this. Then, given any combination of i,j,q,p, it returns the maximum.
Be sure to preallocate
matrix_with_results = zeros(size(resultsMat,1),size(resultsMat,2),size(resultsMat,1),size(resultsMat,2))
for speed.
Two notes:
Don't use i or j as indices/variables names, as they denote the imaginary unit. Using those can easily lead to hard to debug errors.
Initialise matrix_with_results, i.e. tell MATLAB how large it will be before going into the loop. Otherwise, MATLAB will have to increase its size every iteration,which is very slow. This is called preallocation.
Related
I have 2 nested loops which do the following:
Get two rows of a matrix
Check if indices meet a condition or not
If they do: calculate xcorr between the two rows and put it into new vector
Find the index of the maximum value of sub vector and replace element of LAG matrix with this value
I dont know how I can speed this code up by vectorizing or otherwise.
b=size(data,1);
F=size(data,2);
LAG= zeros(b,b);
for i=1:b
for j=1:b
if j>i
x=data(i,:);
y=data(j,:);
d=xcorr(x,y);
d=d(:,F:(2*F)-1);
[M,I] = max(d);
LAG(i,j)=I-1;
d=xcorr(y,x);
d=d(:,F:(2*F)-1);
[M,I] = max(d);
LAG(j,i)=I-1;
end
end
end
First, a note on floating point precision...
You mention in a comment that your data contains the integers 0, 1, and 2. You would therefore expect a cross-correlation to give integer results. However, since the calculation is being done in double-precision, there appears to be some floating-point error introduced. This error can cause the results to be ever so slightly larger or smaller than integer values.
Since your calculations involve looking for the location of the maxima, then you could get slightly different results if there are repeated maximal integer values with added precision errors. For example, let's say you expect the value 10 to be the maximum and appear in indices 2 and 4 of a vector d. You might calculate d one way and get d(2) = 10 and d(4) = 10.00000000000001, with some added precision error. The maximum would therefore be located in index 4. If you use a different method to calculate d, you might get d(2) = 10 and d(4) = 9.99999999999999, with the error going in the opposite direction, causing the maximum to be located in index 2.
The solution? Round your cross-correlation data first:
d = round(xcorr(x, y));
This will eliminate the floating-point errors and give you the integer results you expect.
Now, on to the actual solutions...
Solution 1: Non-loop option
You can pass a matrix to xcorr and it will perform the cross-correlation for every pairwise combination of columns. Using this, you can forego your loops altogether like so:
d = round(xcorr(data.'));
[~, I] = max(d(F:(2*F)-1,:), [], 1);
LAG = reshape(I-1, b, b).';
Solution 2: Improved loop option
There are limits to how large data can be for the above solution, since it will produce large intermediate and output variables that can exceed the maximum array size available. In such a case for loops may be unavoidable, but you can improve upon the for-loop solution above. Specifically, you can compute the cross-correlation once for a pair (x, y), then just flip the result for the pair (y, x):
% Loop over rows:
for row = 1:b
% Loop over upper matrix triangle:
for col = (row+1):b
% Cross-correlation for upper triangle:
d = round(xcorr(data(row, :), data(col, :)));
[~, I] = max(d(:, F:(2*F)-1));
LAG(row, col) = I-1;
% Cross-correlation for lower triangle:
d = fliplr(d);
[~, I] = max(d(:, F:(2*F)-1));
LAG(col, row) = I-1;
end
end
I'm currently working on creating a histogram of Altitudes at which a type of atmospheric instability happens. To be specific, it is when the values of what we call, N^2 is less than zero. This is where the problem comes in. I am trying to plot the occurrence frequency against the altitudes.
load /data/matlabst/DavidBloom/N_square_Ri_number_2005.mat
N_square(N_square > 0) = 0;
N_square = abs(N_square);
k = (1:87);
H = 7.5;
p0 = 101325;
nbins = (500);
N_square(N_square==0)=[];
Alt = zeros(1,578594);
PresNew = squeeze(N_square(:,:,k,:));
for lati = 1:32
for long = 1:64
for t = 1:1460
for k = 1:87
Alt(1,:) = -log((PresNew)/p0)*H;
end
end
end
end
So, let me explain what I am doing. I'm loading a file with all these different variables. Link To Image This shows the different variables it displays. Next, I take the 4-D matrix N_square and I filter all values greater than zero to equal 0. Then I take the absolute value of the leftover negative values. I then define several variables and move on to the next filtering.
(N_square(N_square==0)=[];
The goal of this one was give just discard all values of N_square that were 0. I think this is where the problem begins. Jumping down to the for loop, I am then taking the 3rd dimension of N_square and converting pressure to altitude.
My concern is that when I run this, PresNew = squeeze(N_square(:,:,k,:)); is giving me the error.
Error in PlottingN_2 (line 10)
PresNew = squeeze(N_square(:,:,k,:));
And I have no idea why.
Any thoughts or suggestions on how I could avoid this catastrophe and make my code simpler? Thanks.
When you remove random elements from a multi-dimensional array, they are removed but it can no longer be a valid multi-dimensional array because it has holes in it. Because of this, MATLAB will collapse the result into a vector, and you can't index into the third dimension of a vector like you're trying.
data = magic(3);
% 8 1 6
% 3 5 7
% 4 9 2
% Remove all values < 2
data(data < 2) = []
% 8 3 4 5 9 6 7 2
data(2,3)
% Index exceeds matrix dimensions.
The solution is to remove the 0 values after your indexing (i.e. within your loop).
Alt = zeros(1,578594);
for lati = 1:32
for long = 1:64
for t = 1:1460
for k = 1:87
% Index into 4D matrix
PresNew = N_square(:,:,k,:);
% NOW remove the 0 values
PresNew(PresNew == 0) = [];
Alt(1,:) = -log((PresNew)/p0)*H;
end
end
end
end
Is there any way in Matlab to generate a 5000 x 1000 matrix of random numbers in which:
MM = betarnd(A,B,1,1000);
but A and B are vectors (1 x 5000). I get the following error message:
??? Error using ==> betarnd at 29
Size information is inconsistent.
I want to avoid a loop like the following one:
for ii = 1 : 1000
MM(:,ii) = betarnd(A,B);
end
Thanks!
You can repeat A and B (vectors of size 1x5000) to obtain matrices of size 1000x5000 in which all rows are equal, and use those matrices as inputs to betarnd. That way you get a result of size 1000x5000 in which column k contains 1000 random values with parameters A(k) and B(k).
The reason is that, according to the documentation (emphasis mine):
R = betarnd(A,B) returns an array of random numbers chosen from the
beta distribution with parameters A and B. The size of R is the common size of A and B if both are arrays.
So, use
MM = betarnd(repmat(A(:).',1000,1), repmat(B(:).',1000,1));
I need to calculate the frequency of each value in another vector in MATLAB.
I can use something like
for i=1:length(pdata)
gt(i)=length(find(pf_test(:,1)==pdata(i,1)));
end
But I prefer not to use loop because my dataset is quite large. Is there anything like histc (which is used to find the frequency of values in one vector) to find the frequency of one vector value in another vector?
If your values are only integers, you could do the following:
range = min(pf_test):max(pf_test);
count = histc(pf_test,range);
gt = count(ismember(range,a));
gt(~ismember(unique(a),b)) = 0;
If you can't guarantee that the values are integers, it's a bit more complicated. One possible method of it would be the following:
%restrict yourself to values that appear in the second vector
filter = ismember(pf_test,pdata);
% sort your first vector (ignore this if it is already sorted)
spf_test = sort(pf_test);
% Find the first and last occurrence of each element
[~,last] = unique(spf_test(filter));
[~,first] = unique(spf_test(filter),'first');
% Initialise gt
gt = zeros(length(pf_test));
% Fill gt
gt(filter) = (last-first)+1;
EDIT: Note that I may have got the vectors the wrong way around - if this doesn't work as expected, switch pf_test and pdata. It wasn't immediately clear to me which was which.
You mention histc. Why are you not using it (in its version with two input parameters)?
>> pdata = [1 1 3 2 3 1 4 4 5];
>> pf_test = 1:6;
>> histc(pdata,pf_test)
ans =
3 1 2 2 1 0
I have a 3x3 matrix, A. I also compute a value, g, as the maximum eigen value of A. I am trying to change the element A(3,3) = 0 for all values from zero to one in 0.10 increments and then update g for each of the values. I'd like all of the other matrix elements to remain the same.
I thought a for loop would be the way to do this, but I do not know how to update only one element in a matrix without storing this update as one increasingly larger matrix. If I call the element at A(3,3) = p (thereby creating a new matrix Atry) I am able (below) to get all of the values from 0 to 1 that I desired. I do not know how to update Atry to get all of the values of g that I desire. The state of the code now will give me the same value of g for all iterations, as expected, as I do not know how to to update Atry with the different values of p to then compute the values for g.
Any suggestions on how to do this or suggestions for jargon or phrases for me to web search would be appreciated.
A = [1 1 1; 2 2 2; 3 3 0];
g = max(eig(A));
% This below is what I attempted to achieve my solution
clear all
p(1) = 0;
Atry = [1 1 1; 2 2 2; 3 3 p];
g(1) = max(eig(Atry));
for i=1:100;
p(i+1) = p(i)+ 0.01;
% this makes a one giant matrix, not many
%Atry(:,i+1) = Atry(:,i);
g(i+1) = max(eig(Atry));
end
This will also accomplish what you want to do:
A = #(x) [1 1 1; 2 2 2; 3 3 x];
p = 0:0.01:1;
g = arrayfun(#(x) eigs(A(x),1), p);
Breakdown:
Define A as an anonymous function. This means that the command A(x) will return your matrix A with the (3,3) element equal to x.
Define all steps you want to take in vector p
Then "loop" through all elements in p by using arrayfun instead of an actual loop.
The function looped over by arrayfun is not max(eig(A)) but eigs(A,1), i.e., the 1 largest eigenvalue. The result will be the same, but the algorithm used by eigs is more suited for your type of problem -- instead of computing all eigenvalues and then only using the maximum one, you only compute the maximum one. Needless to say, this is much faster.
First, you say 0.1 increments in the text of your question, but your code suggests you are actually interested in 0.01 increments? I'm going to operate under the assumption you mean 0.01 increments.
Now, with that out of the way, let me state what I believe you are after given my interpretation of your question. You want to iterate over the matrix A, where for each iteration you increase A(3, 3) by 0.01. Given that you want all values from 0 to 1, this implies 101 iterations. For each iteration, you want to calculate the maximum eigenvalue of A, and store all these eigenvalues in some vector (which I will call gVec). If this is correct, then I believe you just want the following:
% Specify the "Current" A
CurA = [1 1 1; 2 2 2; 3 3 0];
% Pre-allocate the values we want to iterate over for element (3, 3)
A33Vec = (0:0.01:1)';
% Pre-allocate a vector to store the maximum eigenvalues
gVec = NaN * ones(length(A33Vec), 1);
% Loop over A33Vec
for i = 1:1:length(A33Vec)
% Obtain the version of A that we want for the current i
CurA(3, 3) = A33Vec(i);
% Obtain the maximum eigen value of the current A, and store in gVec
gVec(i, 1) = max(eig(CurA));
end
EDIT: Probably best to paste this code into your matlab editor. The stack-overflow automatic text highlighting hasn't done it any favors :-)
EDIT: Go with Rody's solution (+1) - it is much better!